Theorem sspsstri 4101

Description: Two ways of stating trichotomy with respect to inclusion. (Contributed by NM, 12-Aug-2004.)

Assertion

Ref	Expression
sspsstri	⊢ ((𝐴 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐴) ↔ (𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵 ∨ 𝐵 ⊊ 𝐴))

Proof of Theorem sspsstri

Step	Hyp	Ref	Expression
1		or32 922	. 2 ⊢ (((𝐴 ⊊ 𝐵 ∨ 𝐵 ⊊ 𝐴) ∨ 𝐴 = 𝐵) ↔ ((𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵) ∨ 𝐵 ⊊ 𝐴))
2		sspss 4098	. . . 4 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵))
3		sspss 4098	. . . . 5 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐵 ⊊ 𝐴 ∨ 𝐵 = 𝐴))
4		eqcom 2737	. . . . . 6 ⊢ (𝐵 = 𝐴 ↔ 𝐴 = 𝐵)
5	4	orbi2i 909	. . . . 5 ⊢ ((𝐵 ⊊ 𝐴 ∨ 𝐵 = 𝐴) ↔ (𝐵 ⊊ 𝐴 ∨ 𝐴 = 𝐵))
6	3, 5	bitri 274	. . . 4 ⊢ (𝐵 ⊆ 𝐴 ↔ (𝐵 ⊊ 𝐴 ∨ 𝐴 = 𝐵))
7	2, 6	orbi12i 911	. . 3 ⊢ ((𝐴 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐴) ↔ ((𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵) ∨ (𝐵 ⊊ 𝐴 ∨ 𝐴 = 𝐵)))
8		orordir 926	. . 3 ⊢ (((𝐴 ⊊ 𝐵 ∨ 𝐵 ⊊ 𝐴) ∨ 𝐴 = 𝐵) ↔ ((𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵) ∨ (𝐵 ⊊ 𝐴 ∨ 𝐴 = 𝐵)))
9	7, 8	bitr4i 277	. 2 ⊢ ((𝐴 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐴) ↔ ((𝐴 ⊊ 𝐵 ∨ 𝐵 ⊊ 𝐴) ∨ 𝐴 = 𝐵))
10		df-3or 1086	. 2 ⊢ ((𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵 ∨ 𝐵 ⊊ 𝐴) ↔ ((𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵) ∨ 𝐵 ⊊ 𝐴))
11	1, 9, 10	3bitr4i 302	1 ⊢ ((𝐴 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐴) ↔ (𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵 ∨ 𝐵 ⊊ 𝐴))

Syntax hints:   ↔ wb 205   ∨ wo 843   ∨ w3o 1084   = wceq 1539   ⊆ wss 3947   ⊊ wpss 3948

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-ext 2701

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 844  df-3or 1086  df-tru 1542  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2722  df-clel 2808  df-ne 2939  df-v 3474  df-in 3954  df-ss 3964  df-pss 3966

This theorem is referenced by:  ordtri3or  6395  sorpss  7720  sorpssi  7721  funpsstri  35041